\(\int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx\) [487]

Optimal result

Integrand size = 22, antiderivative size = 71 \[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\frac {57}{28} (1-x)^{2/3} (2-x)^{2/3}-\frac {3}{7} (1-x)^{5/3} (2-x)^{2/3}-\frac {99}{28} (1-x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-1+x\right ) \] Output:

57/28*(1-x)^(2/3)*(2-x)^(2/3)-3/7*(1-x)^(5/3)*(2-x)^(2/3)-99/28*(1-x)^(2/3 
)*hypergeom([1/3, 2/3],[5/3],-1+x)
                                                                                    
                                                                                    
 

Mathematica [A] (verified)

Time = 10.02 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.62 \[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\frac {3}{28} (1-x)^{2/3} \left ((2-x)^{2/3} (15+4 x)-33 \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},-1+x\right )\right ) \] Input:

Integrate[x^2/((1 - x)^(1/3)*(2 - x)^(1/3)),x]
 

Output:

(3*(1 - x)^(2/3)*((2 - x)^(2/3)*(15 + 4*x) - 33*Hypergeometric2F1[1/3, 2/3 
, 5/3, -1 + x]))/28
 

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {101, 25, 90, 79}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx\)

\(\Big \downarrow \) 101

\(\displaystyle \frac {3}{7} \int -\frac {2-5 x}{\sqrt [3]{1-x} \sqrt [3]{2-x}}dx+\frac {3}{7} (1-x)^{2/3} (2-x)^{2/3} x\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {3}{7} (1-x)^{2/3} (2-x)^{2/3} x-\frac {3}{7} \int \frac {2-5 x}{\sqrt [3]{1-x} \sqrt [3]{2-x}}dx\)

\(\Big \downarrow \) 90

\(\displaystyle \frac {3}{7} (1-x)^{2/3} (2-x)^{2/3} x-\frac {3}{7} \left (-\frac {11}{2} \int \frac {1}{\sqrt [3]{1-x} \sqrt [3]{2-x}}dx-\frac {15}{4} (1-x)^{2/3} (2-x)^{2/3}\right )\)

\(\Big \downarrow \) 79

\(\displaystyle \frac {3}{7} (1-x)^{2/3} (2-x)^{2/3} x-\frac {3}{7} \left (\frac {33}{4} (1-x)^{2/3} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {2}{3},\frac {5}{3},x-1\right )-\frac {15}{4} (1-x)^{2/3} (2-x)^{2/3}\right )\)

Input:

Int[x^2/((1 - x)^(1/3)*(2 - x)^(1/3)),x]
 

Output:

(3*(1 - x)^(2/3)*(2 - x)^(2/3)*x)/7 - (3*((-15*(1 - x)^(2/3)*(2 - x)^(2/3) 
)/4 + (33*(1 - x)^(2/3)*Hypergeometric2F1[1/3, 2/3, 5/3, -1 + x])/4))/7
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 79

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( 
a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 
, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] 
&&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] 
 ||  !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
 

rule 90

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p 
_.), x_] :> Simp[b*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + p + 2))), 
 x] + Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(d*f*(n + p 
+ 2))   Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, 
p}, x] && NeQ[n + p + 2, 0]
 

rule 101

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( 
p_), x_] :> Simp[b*(a + b*x)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d*f*(n + 
 p + 3))), x] + Simp[1/(d*f*(n + p + 3))   Int[(c + d*x)^n*(e + f*x)^p*Simp 
[a^2*d*f*(n + p + 3) - b*(b*c*e + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f 
*(n + p + 4) - b*(d*e*(n + 2) + c*f*(p + 2)))*x, x], x], x] /; FreeQ[{a, b, 
 c, d, e, f, n, p}, x] && NeQ[n + p + 3, 0]
 

Maple [F]

\[\int \frac {x^{2}}{\left (1-x \right )^{\frac {1}{3}} \left (-x +2\right )^{\frac {1}{3}}}d x\]

Input:

int(x^2/(1-x)^(1/3)/(-x+2)^(1/3),x)
 

Output:

int(x^2/(1-x)^(1/3)/(-x+2)^(1/3),x)
 

Fricas [F]

\[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int { \frac {x^{2}}{{\left (-x + 2\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2/(1-x)^(1/3)/(2-x)^(1/3),x, algorithm="fricas")
 

Output:

integral(x^2*(-x + 2)^(2/3)*(-x + 1)^(2/3)/(x^2 - 3*x + 2), x)
 

Sympy [F]

\[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int \frac {x^{2}}{\sqrt [3]{1 - x} \sqrt [3]{2 - x}}\, dx \] Input:

integrate(x**2/(1-x)**(1/3)/(2-x)**(1/3),x)
 

Output:

Integral(x**2/((1 - x)**(1/3)*(2 - x)**(1/3)), x)
 

Maxima [F]

\[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int { \frac {x^{2}}{{\left (-x + 2\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2/(1-x)^(1/3)/(2-x)^(1/3),x, algorithm="maxima")
 

Output:

integrate(x^2/((-x + 2)^(1/3)*(-x + 1)^(1/3)), x)
 

Giac [F]

\[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int { \frac {x^{2}}{{\left (-x + 2\right )}^{\frac {1}{3}} {\left (-x + 1\right )}^{\frac {1}{3}}} \,d x } \] Input:

integrate(x^2/(1-x)^(1/3)/(2-x)^(1/3),x, algorithm="giac")
 

Output:

integrate(x^2/((-x + 2)^(1/3)*(-x + 1)^(1/3)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int \frac {x^2}{{\left (1-x\right )}^{1/3}\,{\left (2-x\right )}^{1/3}} \,d x \] Input:

int(x^2/((1 - x)^(1/3)*(2 - x)^(1/3)),x)
 

Output:

int(x^2/((1 - x)^(1/3)*(2 - x)^(1/3)), x)
 

Reduce [F]

\[ \int \frac {x^2}{\sqrt [3]{1-x} \sqrt [3]{2-x}} \, dx=\int \frac {x^{2}}{\left (1-x \right )^{\frac {1}{3}} \left (-x +2\right )^{\frac {1}{3}}}d x \] Input:

int(x^2/(1-x)^(1/3)/(2-x)^(1/3),x)
 

Output:

int(x**2/(( - x + 1)**(1/3)*( - x + 2)**(1/3)),x)